Study guide1(1)

CHAPTER 1
Computer
Networks and
the Internet
1
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Most Important Ideas and Concepts from Chapter 1
Nuts and bolts of computer networks. Computer networks consist of end systems,
packet switches, and communication links. End systems—also called hosts—in-
clude desktop PCs, laptops, hand-held network devices (including cell phones,
PDAs, and BlackBerries), sensors, and servers (such as Web and mail servers).
Just as cities are interconnected by a network of roads and intersections, end sys-
tems of a computer network are interconnected by a network of communication
links and packet switches. Communication links can be wired or wireless.
Distributed applications. A computer network enables distributed applications.
A distributed application runs on end systems and exchanges data via the com-
puter network. Distributed applications include Web surfing, e-mail, instant mes-
saging, Internet phone, distributed games, peer-to-peer file sharing, television
distribution, and video conferencing. New distributed applications continue to be
invented and deployed on the Internet.
Packet switching. When one end system sends data to another end system, the
sending end system breaks the data into chunks, called packets. Similar to the
process of delivering post-office mail, the Internet transports each packet sepa-
rately, routing a packet to its destination using a destination address that is writ-
ten into the packet. When a packet switch receives a packet, it uses the packet’s
destination address to determine on which link it should forward the packet. Thus,
a packet switch performs “packet switching,” forwarding incoming packets to out-
going links packet by packet. Also, packet switches typically “store and forward”
packets—that is, before a switch begins to forward a packet on an outgoing link,
first it receives and stores the entire packet.
Protocol. A protocol defines the format and order of messages exchanged between
two or more communication entities, as well as the actions taken on the transmission
and/or receipt of a message or other event. Computer networks make extensive use
of protocols. Figure 1.2 (page 7 in your textbook) provides an analogy between a
human protocol and a computer network protocol for messages exchanged between
a Web browser and a Web server. In this example, first, the Web browser sends an in-
troductory message to the server; next, the server responds with its own introducto-
ry message; then, the browser sends another message, requesting a specific Web page;
and finally, the server sends a last message, which includes the requested Web page.
Circuit-switching. Computer networks constitute one major class of communication
networks.Another major class of communication networks is traditional digital tele-
phone networks. Traditional digital telephone networks do not use packet switching
to move data from a source to a destination; they use a technique known as circuit
switching. In circuit switching, before transmitting data between two end systems, the
network establishes a dedicated end-to-end connection between the end systems and
reserves bandwidth in each link along the connection. The reserved connection band-
width is “wasted” whenever the end systems are not sending data.
2 STUDY COMPANION FOR COMPUTER NETWORKING, THIRD EDITION
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Physical media and access networks. The communication links in a computer net-
work may have different physical media types. Dial-up modem links, DSL, and
most Ethernet links are made of copper wire. Cable links are made of coaxial cable.
Long-haul Internet backbone links are made of fiber optics. In addition to these
wired links, there is a plethora of wireless links, including Wi-Fi, Bluetooth®
, and
satellite. An access link is a link that connects the end system to the Internet. Ac-
cess links can be copper wire, coaxial cable, fiber optics or wireless. A tremendous
variety of media types can be found on the Internet.
Network of networks. The Internet consists of many interconnected networks, each
of which is called an Internet Service Provider (ISP). Each ISP is a network of pack-
et switches and communication links. Thus, the Internet is a network of networks.
ISPs are roughly organized in a hierarchy. ISPs at the bottom of the hierarchy access
ISPs such as residential ISPs, university ISPs, and enterprise ISPs. ISPs at the top of
the hierarchy are called tier-1 ISPs and typically include long-haul intra- and inter-
continental fiber links. Tier-n ISPs provide service—for a price—to tier-(n+1) ISPs.
Each ISP is independently managed. However, ISPs employ a common protocol
suite called the Internet Protocol, which is better known as IP.
Transmission and propagation delays. Transmission and propagation delays
play a critical role in the performance of many distributed applications. Perhaps the
best way to understand transmission and propagation delays and their differences
is to work with the Transmission versus Propagation delay applet on the textbook’s
Web site. The propagation delay over a link is the time it takes a bit to travel from
one end of the link to the other. It is equal to the length of the link divided by the
propagation speed of the link’s physical medium. The transmission delay is a quan-
tity that relates to packets and not bits. The transmission delay for a link is equal
to the number of bits in the packet divided by the transmission rate of the link. It
is the amount of time it takes to push the packet onto the link. Once a bit is pushed
onto a link it needs to propagate to the other end. The total delay across a link is
equal to the sum of the transmission delay and the propagation delay.
Queuing delay and packet loss. Many packets can arrive at a packet switch rough-
ly at the same time. If these packets need to be forwarded on the same outbound
link, all but one will have to “queue,” that is, wait to be transmitted. This waiting
introduces a queuing delay. Furthermore, if the queue of packets becomes very
large, the packet switch’s buffer may become exhausted, causing packets to be
dropped or “lost.” Queuing delay and packet loss can severely impact the perfor-
mance of an application.
Protocol layers. A typical computer network makes use of many protocols—easily
hundreds. To deal with this complexity, the protocols are organized into layers. These
protocol layers are arranged in a “stack.” For example, the Internet organizes its pro-
tocols into five layers—namely, from top to bottom: application layer, transport layer,
network layer, link layer, and physical layer. The protocols of layer n use the services
provided by the protocols at the layer (the layer below). This abstract concept,n - 1
CHAPTER 1 • COMPUTER NETWORKS AND THE INTERNET 3
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often difficult to grasp at first, will become clearer as we delve into the different pro-
tocol layers.
With the application-layer as the highest layer in the protocol stack, one can say
that all other layers in the stack exist only to provide services to the application.
Indeed this is the case, as applications are the raison d’être for computer networks.
Without networked applications, there would be no need for computer networks
in the first place!
Encapsulation. When the sender-side application-layer process passes an appli-
cation-level data unit (an application message) to the transport layer, that message
becomes the payload of the transport-layer segment, which also contains addi-
tional transport-layer header information, for example, information that will allow
the transport layer at the receiver side to deliver a message to the correct receiv-
er-side application. Conceptually, one can think of the transport-layer segment as
an envelope with some information on the outside of the envelope (the segment’s
header fields) and the application-layer payload as a message within the envelope.
The transport layer passes the transport-layer segment to the network layer. The seg-
ment becomes the payload of the network-layer datagram, which has additional
fields used by the network layer (for example, the address of the receiver). Con-
ceptually, one can think of the transport-layer segment as an envelope within an
envelope, with some information on the outside of the network-layer envelope.
Finally, the network layer datagram is passed to the link layer, which encapsulates
the datagram within a link-layer frame.
Continuing with our envelope analogy, a receiver-side protocol at layer n will look
at the header information on the envelope. The protocol may pass the envelope back
to the lower layer (for example, for forwarding to another node), or open the en-
velope and extract the upper-layer payload, and pass that upper-layer envelope up
to layer Like layering, the concept of encapsulation can be difficult to grasp
at first. However, the technique is used so often throughout the textbook, that it will
soon become second nature.
n + 1.
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Review Questions
This section provides additional study questions. Answers to each question are pro-
vided in the next section.
1. Packets in the Internet. Suppose end system A wants to send a large file to
end system B. On a very high level, describe how end system A creates pack-
ets from the file. When one of these packets arrives to a packet switch, what
information in the packet does the switch use to determine the link onto
which the packet is forwarded? Why is packet switching analogous to dri-
ving from one city to another and asking directions along the way?
2. Protocol. A protocol defines the format and order of messages exchanged
between two or more communication entities, as well as the actions taken on
the transmission and/or receipt of a message or other event. Describe the
protocol for downloading a Web page, as shown in Figure 1.2 (page 7 in
your textbook).
3. Circuit switching. What are the two most common multiplexing techniques
used in circuit switching?
4. Circuit switching versus packet switching.
a. Suppose that all of the network sources send data at a constant bit rate.
Would packet-switching or circuit-switching be more desirable in this
case? Why?
b. Suppose that all of the network sources are bursty—that they only occa-
sionally have data to send. Would packet-switching or circuit switching
be more desirable in this case? Why?
5. More on circuit switching. Suppose users share a 1 Mbps link. Also sup-
pose each user requires 500 kbps when transmitting, but each user transmits
only 10 percent of the time. (See the discussion on Packet Switching Versus
Circuit Switching in Section 1.3 in your textbook.)
a. When circuit switching is used, how many users can be supported?
b. For the remainder of this question, suppose packet switching is used.
Why will there be essentially no queuing delay before the link if two or
fewer users transmit simultaneously? Why will there be a queuing delay
if three users transmit simultaneously?
c. Find the probability that a given user is transmitting.
d. Suppose there are three users. Find the probability that at any given time,
all three users are transmitting simultaneously. Find the fraction of time
during which the queue grows.
6. Network of networks. List four characteristics of a tier-1 ISP.
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7. More on network of networks. Discuss the economic relationship among
tier-1 ISPs. Discuss the economic relationship between tier-1 and tier-2 ISPs.
8. Propagation delay. How long does it take a packet of length 1,000 bytes to
propagate over a link of distance 5,000 km, propagation speed
and transmission rate 1 Mbps? Generally, how long does it
take a packet of length L to propagate over a link of distance d, propagation
speed s, and transmission rate R bps? Does this delay depend on packet
length? Does this delay depend on transmission rate?
9. Transmission delay. How long does it take to transmit a packet of length
1,000 bytes a link of distance 5,000 km, propagation speed
and transmission rate 1 Mbps? Generally, how long does it take to transmit a
packet of length L over a link of distance d, propagation speed s, and trans-
mission rate R bps? Does this delay depend on the length of the link? Does
this delay depend on the propagation speed of the link?
10. Transmission delay and propagation delay. Consider two packet switches
directly connected by a link of 5000 km, propagation speed
and transmission rate 1 Mbps. How long does it take to move a packet of
length 1,000 bytes from one packet switch to the other packet switch? Gen-
erally, how long does it take to move a packet of length L over a link of dis-
tance d, propagation speed s, and transmission rate R bps?
11. Delays with multiple links. Consider a packet of length L which begins at
end system A, travels over one link to a packet switch, and travels from the
packet switch over a second link to a destination end system. Let and
denote the length, propagation speed, and transmission rate of link i, for
The packet switch delays each packet by Assuming no queu-
ing delays, in terms of and L, what is the total end-to-end
delay for the packet? Suppose the packet is 1,000 bytes, the propagation
speed on both links is the transmission rates of both links is 1
Mbps, the packet switch processing delay is 1 msec, the length of the first
link is 4,000 km, and the length of the last link is 1,000 km. For these values,
what is the end-to-end delay?
12. Store and forwarding. In Question 11, suppose and
Furthermore, suppose the packet switch does not store-and-for-
ward packets but instead immediately transmits each bit it receives before
waiting for the packet to arrive. What is the end-to-end delay?
13. Queuing delay. A packet switch receives a packet and determines the out-
bound link to which the packet should be forwarded. At packet arrival, one
other packet is half transmitted on this outbound link and three other packets
are waiting to be transmitted. Packets are transmitted in order of arrival. Sup-
pose all packets are 1,000 bytes and the link rate is 1 Mbps. What is the queu-
ing delay for the packet? Generally, what is the queuing delay when all packets
dproc = 0.
R1 = R2 = R
2.5 * 108
m/s,
di, si, Ri, (i = 1, 2)
dproc.i = 1, 2.
Ridi, si
2.5 * 108
m/s,
2.5 * 108
m/s,
2.5 * 108
m/s,
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have length L, the transmission rate is R, x bits of the currently transmitted
packet have been transmitted, and N packets are already in the queue?
14. Average queuing delay. Suppose N packets arrive simultaneously to a link
at which no packets are currently being transmitted or queued. Each packet
is of length L and the link has a transmission rate R. What is the average
queuing delay for the N packets?
15. Queuing delay with random arrivals. Consider a link of rate Mbps.
Suppose packets of size bytes arrive “randomly” to the link at a
rate of 1 packet/sec. Let denote the traffic intensity. Suppose the
average queuing delays at the input to this link can be modeled as
as long as The average total delay is the queuing
delay plus the transmission delay. Find the average queuing delay and aver-
age total delay for and 99 packets/sec.
16. Traceroute. Consider the traceroute output on page 44 of the textbook:
a = 30, 60, 90,
I 6 1.[I/(1 - I)] * [L/R]
I = La>R
L = 1,250
R = 1
1 cs-gw (128.119.240.254) 1.009 ms
2 128.119.3.154 (128.119.3.154) 0.931 ms
3 border4-rt-gi-1-3.gw.umass.edu (128.119.2.194) 1.032 ms
4 acr1-ge-2-1-0.Boston.cw.net (208.172.51.129) 10.006 ms
5 agr4-loopback.NewYork.cw.net (206.24.194.104) 12.272 ms
6 acr2-loopback.NewYork.cw.net (206.24.194.62) 13.225 ms
7 pos10-2.core2.NewYork1.Level3.net (209.244.160.133) 12.218 ms
8 gige9-1-52.hsipaccess1.NewYork1.Level3.net (64.159.17.39) 13.081 ms
9 p0-0.polyu.bbnplanet.net (4.25.109.122) 12.716 ms
10 cis.poly.edu (128.238.32.126) 14.080 ms
Here, we display only the first round-trip delay (rather than three delays) for
each router. Explain why the delays sometimes decrease from one router to
the next. Which link has the largest delay? Which link has the second largest
delay? Where are these links located?
17. Circuit switching. Consider sending a packet of F bits over a path of Q
links. Each link transmits at R bps. The network is lightly loaded so that
there are no queuing delays. Propagation delay is also negligible.
a. Suppose the network is a packet-switched datagram network and a con-
nection-oriented service is used. Suppose each packet has bits of
header where Assuming setup time, how long does it take
to send the packet?
b. Suppose that the network is a circuit-switched network. Furthermore,
suppose that the transmission rate of the circuit between source and des-
tination is bps. Assuming setup time and no bits of header ap-
pended to the packet, how long does it take to send the packet?
tsR/24
ts0 6 h 6 1.
h * F
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c. When is the delay longer for packet switching than for circuit switching
assuming Interpret your result.
18. Protocol layers.
a. What are the five protocol layers, from top to bottom, in the Internet?
b. For each of the five layers, what is the name of the packets processed at
the layer?
c. An end-system processes up to which layer?
d. A router processes up to which layer?
e. A link-layer switch processes up to which layer?
h = 0.5?
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Answers to Review Questions
1. End system A breaks the large file into chunks. To each chunk it adds a header,
generating multiple packets from the file. The header in each packet includes
the address of the destination: end system B. The packet switch uses the desti-
nation address to determine the outgoing link. Asking which road to take is
analogous to a packet asking which outgoing link it should be forwarded on,
given the packet’s address. (The driving analogy is discussed on page 23 of
your textbook.)
2. The protocol specifies that four messages are to be sent, in a specific order:
first a TCP connection request message from client to server; then a TCP con-
nection reply message from server to client; then an HTTP request for a specif-
ic Web object from client to server; and finally, a response message from
server to client, including the requested object. The protocol also specifies the
specific format of each of these messages. Note: This example is for illustra-
tive purposes; it is not completely accurate, as we will see in Chapter 2.
3. Frequency division multiplexing, which partitions the bandwidth in a link
using frequency bands, such as in FM radio; and time division multiplexing,
which partitions time in a link with a revolving frame, giving each connec-
tion the same slot in the revolving frame.
4. a. Circuit-switching is more desirable here because there are no statistical
multiplexing gains to be had, and by using circuits, each connection will
get a constant amount of bandwidth that matches its CBR rate. On the
other hand, circuit-switching has more overhead in terms of signaling
needed to set up the call, so there is an argument that packet-switching is
preferable here since there is no call setup overhead with packet-switch-
ing. If this were an exam question either answer would be correct (as
long as you provide the correct reasoning!).
b. Packet-switching is better here because there are statistical multiplexing
gains—when a source does not have data to send, it will not be allocated
bandwidth (it would be unused when the source had nothing to send). With
packet-switching, this bandwidth is available for use by other sources.
5. a. Two users, each getting a dedicated 500 kbps.
b. If two users transmit simultaneously, the input rate is a constant 1 Mbps.
Since the link is also 1 Mbps, there will be no queuing delay. If three
users transmit simultaneously, then the input rate is a constant 1.5 Mbps.
Since the maximum output rate is 1 Mbps, bits will build up in a queue at
a rate of 0.5 Mbps. After one second, the queue length will be approxi-
mately 500,000 bits.
c. Since each user is transmitting 10 percent of the time, the probability that
a user is transmitting at any instant is 0.1.
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d. Assuming that all three users are transmitting independently, the proba-
bility that all of them are transmitting simultaneously is
Since the queue only grows when all three users are transmitting, the
fraction of time during which the queue grows (which is equal to the
probability that all three users are transmitting simultaneously) is .001.
6. • Link speeds are high, typically at gigabit-per-second rates
• Direct connections to all other tier-1 ISPs
• Connected to a large number of tier-2 ISPs
• International coverage
7. There is at least one direct link connection between each pair of tier-1 ISPs.
Typically, tier-1 ISPs do not charge each other for the traffic sent between
them. However, if a tier-2 ISP is directly connected to a tier-1 ISP, the tier-1
ISP charges the tier-2 ISP for the traffic that is sent between the ISPs, there-
by forming a provider/customer service relationship.
8. 20 msec; no; no.
9. 8 msec; no; no.
10. The delay is the sum of the transmission delay and propagation delay. More
specifically, the total delay is the sum of the time for the sending packet
switch to push the packet onto the link and the time for the last bit of the
packet to propagate across the link. From Questions 8 and 9, we have 28
msec;
11. The first end system requires to transmit the packet onto the first link;
the packet propagates over the first link in the packet switch adds a
processing delay of after receiving the entire packet, the packet switch
requires to transmit the packet onto the second link; the packet propa-
gates over the second link in Adding these five delays gives
To answer the second question, we simply plug the values into the equation
to get
12. Because bits are immediately transmitted, the packet switch does not introduce
any delay; in particular, it does not introduce a transmission delay. Thus,
For the values in Question 11, we get .
13. The arriving packet must first wait for the link to transmit 3,500 bytes or
28,000 bits. Since these bits are transmitted at 1 Mbps, the queuing delay is
28 msec. Generally, the queuing delay is
14. The queuing delay is 0 for the first transmitted packet 0, for the second
transmitted packet, and, generally, for the nth transmitted pack-
et. Thus, the average delay for the N packets is
(n - 1)L>R
L>R
[NL + (L - x)]>R.
8 + 16 + 4 = 28 msec
dend-end = L>R + d1>s1 + d2>s2 .
8 + 8 + 16 + 4 + 1 = 37 msec.
dend-end = L>R1 + L>R2 + d1>s1 + d2 >s2 + dproc.
d2>s2.
L>R2
dproc;
d1>s1;
L>R1
L>R + d>s.
L>R;
d>s;
(0.1)3
= .001.
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Note that here we used the well-known fact that
15. Thus, we have:
The transmission delay for each case is The total delay is
the queuing delay plus the transmission delay. For the four cases, this is 14.3
msec, 25 msec, 100 msec, and 1 sec.
16. The roundtrip to router 8 is 13.1 msec and the roundtrip delay to router 9 is
12.7. Two separate roundtrip probes were sent to routers 8 and 9 at two (slight-
ly) different times. During these times, congestion in the links along the path
changed. Apparently there was more congestion in at least one of the first eight
links during the first probe than in the second probe; hence, the roundtrip delay
to router 8 is greater than the roundtrip delay to router 9. The two largest de-
lays occur between routers 3 and 4 and between routers 4 and 5. Routers 3 and
4 appear to be located in Amherst Massachusetts (UMass) and Boston. Routers
4 and 5 appear to be located in Boston and New York City.
17. a. The time required to transmit the packet over one link is
The time required to transmit the packet over Q links is
Thus, the total delay for packet switching is
b. For circuit switching, bits are not “store and forwarded” before each link.
Thus, there is only one transmission delay of The
total delay is
c. The delay is longer for packet switching when
or equivalently when Thus,
if there are more than 16 links, packet switching has a larger delay, due
to the store and forwarding. If there are fewer than 16 links, circuit
switching has a larger delay, due to its reduced transmission rate
18. a. application, transport, network, link, physical
b. message, segment, datagram, frame, packet
c. an end-system processes up through the application layer
d. a router processes up through the network layer
e. a link-layer switch processes up through the link layer
(R>24).
Q 7 16.Q(1 + h)F>R + ts 7 24F>R + ts
24F>R + ts.
F>(R>24) = 24F>R.
Q(1 + h)F>R + ts.
Q(1 + h)F>R.
(1 + h)F>R.
L>R = 10 msec.
a = 99; I = .99, average queuing delay = (.99>.01) 10 msec = 990 msec
a = 30: I = .3, average queuing delay = (.3>.7) 10msec = 4.3 msec
L>R = 10 msec.
1 + 2 + Á + N = N(N + 1)>2.
+ (N - 1)) = LN(N - 1)>2RN = (N - 1)L>2R
(L>R + 2L>R + Á + (N - 1)L>R)>N = L>RN(1 + 2 + Á
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CHAPTER 2
The Application
Layer
13
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Most Important Ideas and Concepts from Chapter 2
Application-layer protocol. In Chapter 1 we noted that “A protocol defines the
format and the order of messages exchanged between two or more communicat-
ing entities, as well as the actions taken on the transmission and/or receipt of a mes-
sage or other event.” In chapter 2, we have seen how processes send and receive
messages in an application-layer protocol. As a review, identify the messages ex-
changed and actions taken by the following protocols: HTTP, FTP, DNS, SMTP.
Client/server versus peer-to peer. These are the two approaches that we studied
structuring a network application. In the client/server paradigm (see page 75 of
the textbook), a client process requests a service by sending one or more messages
to a server process. The server process implements a service by reading the client
request, performing some action (for example, in the case of an HTTP server, find-
ing a Web page), and sending one or more messages in reply (in the case of HTTP,
returning the requested object). In a peer-to-peer approach, the two ends of the
protocol are equals (as in a telephone call).
Two services provided by the Internet’s transport layer: reliable, congestion-
controlled data transfer (TCP), and unreliable data transfer (UDP). These are
the only services available to an Internet application to transfer data from one process
to another remote process. The Internet transport layer does not provide a minimum
guaranteed transfer rate, or a bound on the delay from source to destination.
HTTP: request/response interaction. The HTTP protocol is a simple applica-
tion-layer protocol. A client (Web browser) makes a request with a GET message,
and a Web server provides a reply (see Figure 2.6 on page 89 in your textbook).
This is a classical client/server approach. Since HTTP uses TCP to provide reliable
transfer of the GET request from client-to-server, and the reply from server-to-
client, a TCP connection must be set up. ATCP setup request is sent from the TCP
in the client to the TCP in the server, with the TCP server replying to the TCP
client. Following this exchange, the HTTP GET message can be sent over the TCP
connection from client-to-server, and the reply received (see Figure 2.7 on page 92
in your textbook). With non-persistent HTTP, a new TCP connection must be set
up each time the client wants to contact the server. With persistent HTTP, multi-
ple HTTP GET messages can be sent over a single TCP connection, resulting in
performance gains from not having to set up a new TCP for each of the HTTP re-
quests beyond the first.
Caching. Caching is the act of saving a local copy of a requested piece of infor-
mation (for example, Web document or DNS translation pair) that is retrieved from
a distant location, so that if the same piece of information is requested again, it can
be retrieved from the local cache, rather than having to retrieve the information
again from the distant location. Caching can improve performance by decreasing
response time (since the local cache is closer to the requesting client) and avoiding
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the use of scare resources (such as the 1.5 Mbps access link shown in Figures 2.11
on page 103 and 2.12 on page 104 in your textbook). Think about ways you use
caching in your every day life—e.g., writing a phone number on a piece of paper
and keeping it in your pocket, rather than looking it up again in a phone book.
DNS: core infrastructure implemented as an application-layer process. The
DNS is an application-layer protocol. The name-to-IP-address translation service
is performed at the DNS servers, just as any application provides a service to a
client via a server. But the DNS service is a very special network service—with-
out it the network would be unable to function. Yet it is implemented in very much
the same way as any other network application.
FTP: separate control and data. Students often ask us why we include an “old”
application such as FTP here. FTP is a nice example of a protocol that separates con-
trol and data messages.As shown in Figure 2.15 (on page 110 in your textbook), con-
trol and data messages are sent over separate TCP connections. This logical and
physical separation of control and data (rather than mixing the two types of messages
in one connection) helps to make the structure of such an application “cleaner.”
TCP sockets: accept(), and the creation of a new socket. A “tricky” thing about
TCP sockets is that a new socket is created when a TCP server returns from an ac-
cept() system call. We call the socket on which the server waits when performing
the accept() as a “welcoming socket.” The socket returned from the accept() is
used to communicate back to the client that connected to the server via the ac-
cept() (see Figure 2.28 on page 150 in your textbook).
UDP sockets: send and pray on the receiving side; datagrams from many
senders on the receiving side. Since UDP provides an unreliable data transfer
service, a sender that sends a datagram via a UDP socket has no idea if the data-
gram is received by the receiver (unless the receiver is programmed to send back
a datagram that acknowledges that the original datagram was received). On the
receiving side, datagrams from many different senders can be received on the same
socket.
Pull versus push. How does one application process get data to or from another
application process? In a pull system (such as the Web), the data receiver must ex-
plicitly request (“pull”) the information. In a push system, the data holder sends
the information to the receiver without the receiver’s explicitly asking for the data
(as in SMTP, when an email is “pushed” from sender to receiver).
Locating information in P2P systems. We identified three ways to locate infor-
mation in a P2P system: query flooding, directory systems, and hybrid systems. All
existing P2P systems use one of these approaches.
CHAPTER 2 • THE APPLICATION LAYER 15
CH01-02_p1-30 6/15/06 4:34 PM Page 15

Review Questions
This section provides additional study questions. Answers to each question are pro-
vided in the next section.
1. Client-server, P2P, or Hybrid? Section 2.1.1 in your textbook discusses
three application architectures: client-server, P2P, and a hybrid of the two.
Classify each of the scenarios below as client-server, P2P, or hybrid, and ex-
plain your answer briefly. Answering these questions may require some Web
surfing.
a. EBay
b. Skype
c. BitTorrent
d. Telnet
e. DNS
2. Services provided by the Internet transport protocols. Indicate whether TCP
or UDP (or both or neither) provide the following services to applications:
a. Reliable data transfer between processes.
b. Minimum data transmission rate between processes.
c. Congestion-controlled data transfer between processes.
d. A guarantee that data will be delivered within a specified amount of time.
e. Preserve application-level message boundaries. That is, when a sender
sends a group of bytes into a socket via a single send operation, that
group of bytes will be delivered as a group in a single receive operation
at the receiving application.
f. Guaranteed in-order delivery of data to the receiver.
3. Fast transactions. Suppose you want to do a transaction from a remote
client to a server as fast as possible, would you use UDP or TCP?
4. Reliable data transfer with UDP. Suppose you use UDP to do a transaction
from a remote client to a server. UDP provides no reliability, but you want
your transaction request to be sent reliably. How could you do is?
5. Timeliness or in-order delivery. Suppose that data is being output at a sen-
sor at 1 sample per second. It is important for the receiver to have the most
recent value of the sensor’s reading, rather than all values (for example, it is
better for the receiver to get a current value, rather than an outdated value).
Would you use TCP or UDP to send this sensor data? Why?
6. HTTP basics. Consider the following string of ASCII characters that were
captured by Ethereal when the browser sent an HTTP GET message (this is the
CH01-02_p1-30 6/15/06 4:34 PM Page 16

actual content of an HTTP GET message). The characters <cr><lf> are car-
riage return and line feed characters (that is, the italicized character string
<cr> in the text below represents the single carriage-return character that was
contained at that point in the HTTP header). Answer the following questions,
indicating where in the HTTP GET message below you find the answers.
GET /cs453/index.html HTTP/1.1<cr><lf>Host: gai
a.cs.umass.edu<cr><lf>User-Agent: Mozilla/5.0 (
Windows;U; Windows NT 5.1; en-US; rv:1.7.2) Gec
ko/20040804 Netscape/7.2 (ax) <cr><lf>Accept:ex
t/xml,application/xml,application/xhtml+xml,text
/html;q=0.9,text/plain;q=0.8,image/png,*/*;q=0.5
<cr><lf>Accept-Language: en-us,en;q=0.5..Accept-
Encoding: zip,deflate<cr><lf>Accept-Charset: ISO
-8859-1,utf-8;q=0.7,*;q=0.7..Keep-Alive: 300<cr>
<lf>Connection:keep-alive<cr><lf><cr><lf>
a. What is the URL of the document requested by the browser? Make sure
you give the hostname and the file name parts of the URL.
b. What version of HTTP is the browser running?
c. Is a Netscape or an Internet Explorer browser making the request?
d. Is the browser requesting a non-persistent or a persistent connection?
e. What is the IP address of the computer on which the browser is running?
7. More HTTP basics. The text below shows the reply sent from the server in
response to HTTPP GET message in Question 6. Answer the following ques-
tions, indicating where in the message below you find the answers.
HTTP/1.1 200 OK<cr><lf>Date: Tue, 07 Mar 2006
12:39:45GMT..Server: Apache/2.0.52 (Fedora)
<cr><lf>Last-Modified: Sat, 10 Dec 2005 18:27:46
GMT<cr><lf>ETag: "526c3-f22-a88a4c80"<cr><lf>Accept-
Ranges: bytes<cr><lf>Content-Length: 3874<cr><lf>Keep-
Alive: timeout=max=100<cr><lf>Connection: Keep-
Alive<cr><lf>Content-Type: text/html; charset=ISO-8859-
1<cr><lf><cr><lf><!doctype html public "-//w3c//dtd html
4.0 transitional//en"><lf><html><lf><head><lf> <meta
http-equiv="Content-Type" content="text/html;
charset=iso-8859-1"><lf> <meta name="GENERATOR"
content="Mozilla/4.79 [en] (Windows NT 5.0; U)
Netscape]"><lf> <title>CMPSCI 453 / 591 / NTU-ST550A
Spring 2005 homepage</title><lf></head><lf> <much more
document text following here (not shown)>
CH01-02_p1-30 6/15/06 4:34 PM Page 17

a. Was the server able to find the document successfully or not?
b. At what time was the document reply provided?
c. When was the document last modified?
d. How many bytes are there in the document being returned?
e. What are the first 5 bytes of the document being returned?
f. Did the server agree to a persistent connection?
8. HTTP Performance. Here, we consider the performance of HTTP, comparing
non-persistent HTTP with persistent HTTP. Suppose the page your browser
wants to download is 100K bits long, and contains 10 embedded images (with
file names img01.jpg, img02.jpg, … img10.jpg), each of which is also 100K bits
long. The page and the 10 images are stored on the same server, which has a 300
msec roundtrip time (RTT) from your browser. We will abstract the network path
between your browser and the Web server as a 100 Mbps link. You can assume
that the time it takes to transmit a GET message into the link is zero, but you
should account for the time it takes to transmit the base file and the embedded
objects into the “link.” This means that the server-to-client “link” has both a 150
msec one-way propagation delay, as well as a transmission delay associated with
it. (Review page 39 in the textbook if you are uncertain about the difference be-
tween transmission delay and propagation delay.) In your answer, be sure to ac-
count for the time needed to set up a TCP connection (1 RTT).
a. Assuming non-persistent HTTP (and assuming no parallel connections
are open between the browser and the server), how long is the response
time—the time from when the user requests the URL to the time when
the page and its embedded images are displayed? Be sure to describe the
various components that contribute to this delay.
b. Again, assume non-persistent HTTP, but now assume that the browser
can open as many parallel TCP connections to the server as it wants.
What is the response time in this case?
c. Now assume persistent HTTP (HTTP1.1). What is the response time, as-
suming no pipelining?
d. Now suppose persistent HTTP with pipelining is used. What is the re-
sponse time?
9. Caching and delays. Consider the networks shown in the figure below. There
are two user machines—m1.a.com and m2.a.com in the network a.com. Sup-
pose the user at m1.a.com types in the URL www.b.com/bigfile.htm into a
browser to retrieve a 1Gbit (1000 Mbit) file from www.b.com.
a. List the sequence of DNS and HTTP messages sent/received from/by
m1.a.com, as well as any other messages that leave/enter the a.com network
that are not directly sent/received by m1.a.com from the point that the URL
is entered into the browser until the file is completely received. Indicate the
CH01-02_p1-30 6/15/06 4:34 PM Page 18

source and destination of each message. You can assume that every HTTP
request by m1.a.com is first directed to the HTTP cache in a.com, that the
cache is initially empty, and that all DNS requests are iterated queries.
b. How long does it take to accomplish the steps you outlined in your an-
swer to the previous question regarding the m1a.com HTTP and DNS
messages. Explain how you arrived at your answer. In answering this
question, you can assume the following:
• The packets containing DNS commands and HTTP commands such as
GET are very small compared to the size of the file. Therefore, their
transmission times (but not their propagation times) can be neglected.
• Propagation delays within the local area networks (LANs) are small
enough to be ignored. The propagation from router R1 to router R2 is
100 msec.
• The one-way propagation delay from anywhere in a.com to any other
site in the Internet (except b.com) is 500 msec.
c. Now assume that machine m2.a.com makes a request to the same URL
that m1.a.com requested. List the sequence of DNS and HTTP messages
sent/received from/by m2.a.com as well as other messages that
leave/enter the a.com network that are not directly sent/received by
m2.a.com from the point that the URL is entered into the browser until
the file is completely received. Indicate the source and destination of
each message. (Hint: be sure to consider caching.)
CH01-02_p1-30 6/15/06 4:34 PM Page 19

d. Now suppose there is no HTTP cache in network a.com. What is the
maximum rate at which machines in a.com can make requests for the file
www.b.com/bigfile.htm while keeping the time from when a request is
made to when it is satisfied non-infinite in the long run?
10. Persistent versus non-persistent TCP connections. Suppose within your
Web browser you click on a link to obtain a Web page. Suppose the IP ad-
dress for the associated URL is cached in your local host, so that a DNS
lookup is not necessary. Denote RTT as the roundtrip time between the local
host and the server containing the Web page. Assume the Web page consists
of a base HTML file and three small images. Assume the transmission times
for all of the objects are negligible in comparison with the RTT. How much
time elapses (in terms of RTTs) from when the user clicks on the link until
the client receives the entire Web page with each of the following?
a. Non-persistent HTTP with no parallel connections
b. Non-persistent HTTP with up to five parallel connections
c. Persistent HTTP with pipelining
11. In-band versus out-of band control. What does it mean when we say that
control messages are “in-band”? What does it mean when we say that con-
trol messages are “out-of-band”? Give an example of a protocol that has in-
band control messages and an example of a protocol that has out-of-band
control messages.
CH01-02_p1-30 6/15/06 4:34 PM Page 20

12. Networking over a slow, short link. Consider a short (10 meter) link, over
which a sender can transmit at a rate of 100 bits/sec in both directions. Sup-
pose that packets containing data are 100Kbits long, and packets containing
only control (for example, ACK or handshaking) are 100 bits long. Assume
that N parallel connections each get 1/N of the link bandwidth.
Now consider the HTTP protocol, and suppose that each downloaded object
is 100Kbits long, and that the initial downloaded object contains 10 refer-
enced objects from the same sender. Would parallel downloads via parallel
instances of non-persistent HTTP make sense in this case? Now consider
persistent HTTP. Do you expect significant gains over the non-persistent
case? Explain your answer.
13. Who is sending this packet? Is the following statement true or false? With
UDP sockets, a server can easily determine the IP address of the client, from
the data returned via a socket read. Answer the same question (true or false?)
for TCP sockets. Briefly explain your answers.
14. Caching. Describe two ways in which caching is used in Web access. De-
scribe one way in which caching is used in DNS.
15. Mail. What is the difference between the MAIL FROM: in SMTP and From:
in the mail message itself?
16. Push versus pull. Consider the following forms of non-Internet data distrib-
ution: FM radio, broadcast TV, and newspapers. Are these “push” or “pull”
systems? Explain. Is the traditional Web browser/server a “push” or “pull”
system? Explain. Are there any non-Internet “pull” systems in your life?
17. Query flooding in P2P networks. Here, we explore the reverse-path routing
of the QueryHit messages in Gnutella. Suppose that Alice issues a Query
message. Furthermore, suppose that Bob receives the Query message (which
may have been forwarded by several intermediate peers) and has a file that
matches the query.
a. Recall that when a peer has a matching file, it sends a QueryHit message
along the reverse path of the corresponding Query message. An alterna-
tive design would be for Bob to establish a direct TCP connection with
Alice and send the QueryHit message over this connection. What are the
advantages and disadvantages of such an alternative design?
b. In the Gnutella protocol, when the peer Alice generates a Query mes-
sage, it inserts a unique ID in the message’s MessageID field. When the
peer Bob has a match, it generates a QueryHit message using the same
MessageID as the Query message. Describe how peers can use the Mes-
sageID field and local routing tables to accomplish reverse-path routing.
c. An alternative approach, which does not use message identifiers, is as fol-
lows. When a query message reaches a peer, before forwarding the mes-
CH01-02_p1-30 6/15/06 4:34 PM Page 21

sage, the peer augments the query message with its IP address. Describe
how peers can use this mechanism to accomplish reverse-path routing.
18. TCP sockets: accept(). What is the purpose of the connection-oriented wel-
coming socket, which the server uses to perform an accept()? Once the ac-
cept() is done, does the server use the welcoming socket to communicate
back to the client? Explain your answer.
19. How many port numbers are in use? Consider incoming TCP and UDP
segments arriving at a server, and suppose that (for example, using Ethereal),
we see that 150 different destination port numbers are being used. The server
acts only as a server (that is, in the socket sense, it does not initiate commu-
nication with any other computers as a client; it only responds to incoming
segments). Can the number of sockets in use at the server be (a) larger than,
(b) equal to, or (c) less than 150? Explain your answer.
20. Socket programming. The skeleton of TCPServer.java is given below. This
server receives a sentence from a client on port 6789, capitalizes the sen-
tence, and sends the sentence back to the client. Answer the following ques-
tions without looking at the code in the textbook:
a. Provide the one line of code that belongs at location (a).
b. Provide the one line of code that belongs at location (b).
import java.io.*;
import java.net.*;
class TCPServer {
public static void main(String argv[]) throws Exception
{
String clientSentence;
String capitalizedSentence;
// (a) Insert line of code
while(true) {
// (b) Insert line of code
BufferedReader inFromClient = new BufferedReader(new
InputStreamReader(connectionSocket.getInputStream()));
DataOutputStream outToClient = new
DataOutputStream(connectionSocket.getOutputStream());
clientSentence =
inFromClient.readLine();
capitalizedSentence =
clientSentence.toUpperCase() + 'n';
outToClient.writeBytes(capitalizedSentence);
}
}
}
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Answers to Review Questions
1. a. EBay is pure client-server application architecture. Ebay is implemented
as a Web server (more accurately, a farm of Web servers) that responds to
Web client (browser) requests using HTTP.
b. When two Skype clients talk to each other, they do so as peers. However,
in order to locate a peer, a Skype client will first contact a directory serv-
er in a client-server manner. Therefore, Skype has a hybrid application
architecture.
c. BitTorrent is a pure P2P application architecture. It is interesting because it
will concurrently download different pieces of a file from different peers.
d. Telnet is a pure client-server application architecture. The client is the
process that contacts the Telnet server (at port 23 to allow remote login
on the remote machine where the Telnet server process is running).
e. DNS is a pure client-server application architecture. The DNS client is
the process that sends the DNS REQUEST message (to port 53 at the
DNS server); the server is the DNS server process that replies with a
DNS REPLY message.
2 a. TCP provides a reliable byte-stream between client and server.
b. Neither
c. TCP
d. Neither
e. UDP preserves message boundaries, while TCP is byte-stream oriented,
and does not preserve message boundaries.
f. TCP will deliver bytes to the receiver in the order in which they were
sent. UDP does not guarantee delivery of message data in the order in
which they were sent.
3. You would use UDP. With UDP, the transaction can be completed in one
roundtrip time (RTT)—the client sends the transaction request into a UDP
socket, and the server sends the reply back to the client’s UDP socket. With
TCP, a minimum of two RTTs are needed—one to set-up the TCP connec-
tion, and another for the client to send the request, and for the server to send
back the reply.
4. You would build reliability into your application. This could be done, for ex-
ample, by having the client re-transmit its request if it doesn’t hear back
from the server within a certain amount of time. We will cover techniques to
provide reliable data transfer in Chapter 3.
5. You would use UDP. With TCP, data (even old data) will be sent again and
again until it is received correctly. Moreover, since data is passed up in order
CH01-02_p1-30 6/15/06 4:34 PM Page 23

by the TCP receiver, newer data will not be delivered to the receiving appli-
cation until all old data has been delivered to the application. With UDP, if
data is lost, newer data will eventually be sent and received, without waiting
for the lost data to be recovered.
6. a. The document request was http://gaia.cs.umass.edu/cs453/index.html.
The Host: field indicates the server’s name and /cs453/index.html
indicates the file name.
b. The browser is running HTTP version 1.1, as indicated just before the
first <cr><lf> pair.
c. A Netscape browser is making the request. The User-agent: field indi-
cates “Mozilla,” which is a nickname for Netscape’s browser.
d. The browser is requesting a persistent connection, as indicated by the
Connection:keep-alive.
e. This is a trick question. This information is not contained in an HTTP
message anywhere. So there is no way to tell this from looking at the ex-
change of HTTP messages alone. One would need information from the
IP datagrams (that carried the TCP segment that carried the HTTP GET
request) to answer this question.
7. a. The status code of 200 and the phrase OK indicate that the server was
able to locate the document successfully.
b. The reply was provided on Tuesday, 07 Mar 2006 12:39:45 Greenwich
Mean Time.
c. The document index.html was last modified on Saturday, 10 Dec 2005
18:27:46 GMT.
d. There are 3874 bytes in the document being returned.
e. The first five bytes of the returned document are: <!doc.
f. The server agreed to a persistent connection, as indicated by the Con-
nection: Keep-Alive field.
8. a. For starters, note that it takes 1 msec to send 100K bits over a 100 Mbps
link. The delays associated with this scenario are:
• 300 msec (RTT) to set up the TCP connection that will be used to re-
quest the base file.
• 150 msec (one way delay) to send the GET message for the base file,
and have the message propagate to the server, plus 1 msec to transmit
the base file, plus 150 msec for the base file to propagate back to the
client (for a total of 301 msec).
• 300 msec (RTT) to set up TCP connection that will be used to request
the img.01.jpg file.
• 150 msec (one way delay) to send the GET message for img01.jpg and
have it propagate to the server, plus 1 msec to transmit the img01.jpg
CH01-02_p1-30 6/15/06 4:34 PM Page 24

file, plus 150 msec for the img01.jpg file to propagate back to the
The last two steps above are repeated for the nine image files img02.jpg
through img10.jpg. The total response time is therefore
or 6.911 seconds.
b. • 300 msec (RTT) to set up the TCP connection that will be used to re-
quest the base file.
• The client now sets up 10 parallel TCP connections. 300 msec (RTT) is
needed to set up the 10 TCP connections (since they are set up in parallel).
• 150 msec (one way delay) to send the 10 GET messages in parallel for
img01.jpg through img10.jpg and have the GET messages propagate to
the server. It will take the server 10 msec to transmit the 10 jpeg files,
plus 150 msec for the last jpeg file to propagate back to the client (for a
total of 310 msec).
Putting this all together, the response time is
seconds.
c. • 300 msec (RTT) to set up the TCP connection that will be used to re-
quest the base file, and 10 images.
• 150 msec (one way delay) to send the GET message for img01.jpg and
have it propagate to the server, plus 1 msec to transmit the img01.jpg
file, plus 150 msec for the img01.jpg file to propagate back to the
The last step above is repeated for the nine image files img02.jpg through
img10.jpg. The total response time is therefore
d. • 300 msec (RTT) to set up the TCP connection that will be used to re-
quest the base file, and 10 images.
300 + 11 * 301 = 3.611 seconds
300 + 301 + 300 + 310 = 1.211
300 + 11 * 601,
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• 150 msec (one way delay) to send the 10 GET messages serially for
img01.jpg through img10.jpg and have the GET messages propagate
to the server (recall that we are assuming the GET message has zero
transmission time). It will take the server 10 msec to transmit the 10
jpeg files, plus 150 msec for the last jpeg file to propagate back to the
The total response time is
9. a. • m1.a.com needs to resolve the name www.b.com to an IP address so it
sends a DNS REQUEST message to its local DNS resolver.
• The local DNS server does not have any information so it contacts a
root DNS server with a REQUEST message.
• The root DNS server returns name of DNS Top Level Domain server
for .com.
• The local DNS server contacts the .com TLD.
• The TLD .com server returns the authoritative name server for b.com.
• The local DNS server contacts the authoritative name server for b.com.
• The authoritative name server for b.com returns the IP address of
www.b1.com.
• The HTTP client sends a HTTP GET message to www.b1.com, which is
redirected by the client browser to the HTTP cache in the a.com network.
• The HTTP cache does not find the requested document in its cache, so
it sends the GET request to www.b.com.
• www.b.com receives the GET request and sends the file from
www.b.com to R2.
• The 1 Gbit file is transmitted over the 1 Mbps link between R2 and R1
to the HTTP cache.
• The 1 Gbit file is sent from the HTTP cache to m1.a.com.
b. Let be the time at which the user enters www.b.com into the
browser
• The HTTP client will send its HTTP GET message to www.b1.com
through the local HTTP cache in a.com. This takes no time given the
assumptions above.
• The HTTP cache does not find the requested document in its cache.
Therefore it must request the document from b.com. Before it can send
a GET request to www.b.com, it must find out the IP address for
www.b.com. To do this, it will have to use the DNS. To resolve the
name www.b.com to an IP address, the Web cache sends a DNS RE-
QUEST message to its local DNS resolver. This takes no time given
the assumptions above.
t = 0
300 + 301 + 310 = 911 msec.
CH01-02_p1-30 6/15/06 4:34 PM Page 26

• The local DNS server does not have any information so it contacts a
root DNS server with a REQUEST message. This takes .5 sec given
the assumptions above. At msec, the root DNS server receives
the message.
• The root DNS server returns the name of the DNS Top Level Domain
server for .com. This takes 500 msec given the assumptions above. At
sec, the local DNS server receives the message and now knows
the address of the .com TLD DNS server.
• The local DNS server contacts the .com TLD DNS server. This takes
500 msec given the assumptions above. At the TLD DNS
server receives the message.
• The TLD .com server returns the authoritative DNS server for b.com.
This takes 500 msec given the assumptions above. At the local
DNS server receives the message.
• The local DNS server contacts the authoritative name server for b.com.
This takes 100 msec given the assumptions above. At the au-
thoritative DNS server for b.com receives the message.
• The authoritative name server for b.com returns the IP address of
www.b1.com. This takes 100 msec, given the assumptions above. At
the local DNS server for a.com receives the message, and re-
turns this message to the HTTP Web cache.
• The Web cache is now ready to request the document from
www.b.com. It takes 200 msec to set up the TCP connection between
the HTTP cache and www.b.com.
• At sec, the Web cache sends the GET request to www.b.com.
It takes 100 msec for the GET request to propagate to www.b.com,
given the assumptions above.
• At sec, www.b.com receives the GET request and immediately
begins sending the file in reply. If we assume that as soon as the first
bit reaches R2, it is forwarded over the link between R2 and R1, then
this transmission delay can be ignored, since the transmission of the
file from www.b.com is pipelined with the transmission of the file be-
tween R2 and R1.
• The 1 Gbit file must be transmitted over the 1 Mbps link between R2
and R1. This takes 1,000 seconds. There is an additional 100 msec
propagation delay. Thus, at secs, the last bit of the file is
received at R1. If we assume that R1 forwards packets to the HTTP
cache as it receives them from R2, then the transmission delay be-
tween R1 and the cache can be ignored since it is pipelined with the
transmission from R2 to R1.
t = 1002.6
t = 2.5
t = 2.4
t = 2.2,
t = 2.1,
t = 2,
t = 1.5,
t = 1
t = 500
CH01-02_p1-30 6/15/06 4:34 PM Page 27

• There is a 1 sec delay to send the 1 Gbps file from R1 to the HTTP cache.
If we assume that as soon as the first few bits of the file arrive at the
router, they are forwarded to the cache, then this delay can be ignored.
• There is a 1 sec delay to send the 1 Gbps file from the HTTP cache to
m1.a.com. If we assume that as soon as the first few bits of the file ar-
rive at the cache, they are forwarded to the m1.a.com, then this delay
can be ignored.
Thus, the total delay is approximately 1002.6 seconds.
c. • The HTTP client at m2.a.com will send its HTTP GET message to
www.b1.com through the local HTTP cache in a.com. This takes no
time given the assumptions above.
• The HTTP cache finds the requested document in its cache, so it sends
a GET request with an If-Modified-Since to www.b.com. This takes
100 msec given the assumptions above. Note that the cache does not
need to contact the DNS, assuming it has cached the IP address associ-
ated with www.b.com.
• www.b.com receives the GET request. The document has not changed,
so www.b.com sends a short HTTP REPLY message to the HTTP
cache in a.com indicating that the cached copy is valid. (This takes
100 msec given the assumptions.)
• There is a 1 sec delay to send the 1 Gbps file from the HTTP cache to
m2.a.com.
Thus, the total delay is: sec.
d. Since it takes approximately 1000 sec to send the file from R2 to R1, the
maximum rate at which requests to send the file from b.com to a.com is
1 request every 1000 seconds, or an arrival rate of .001 requests/sec.
10. a. 2 RTT to get each image:
b. 2 RTT for base, 2 RTT for remaining three images: 4 RTT
c. 2 RTT for base; 1 RTT for remaining three images: 3 RTT
11. When we say that control messages are “in-band,” it means that control mes-
sage and data messages may be interleaved with each other on the same con-
nection. A single message may contain both control information and data.
When we say that control messages are “out-of-band,” it means that control
and data messages are carried on separate connections. HTTP, DNS, and
SMTP have in-band control, while FTP has out-of-band control messages.
12. Parallel download would only share the 100K bandwidth among the 10 con-
nections (each getting just 10K bits/sec) thus, there is no significant advan-
tage here. With persistent HTTP we avoid the SYN and SYNACK exchange,
but that only requires 2 seconds (1 second to send the 100 bit SYN message
4(2 RTT) = 8 RTT
.1 + .1 + 1 = 1.2
CH01-02_p1-30 6/15/06 4:34 PM Page 28

over the 100 bps link, and 1 second to receive the ACK). Given that each ob-
ject takes 101 seconds to send and receive the ACK, the use of pipelining
gives only a 2 percent gain.
13. This is true for UDP since the UDP packet (i.e., the Datagram Packet data
structure returned from the ServerSocket.read( ) call) contains the IP address
of the sender of the UDP packet. This is false for a TCP socket since the
socket only returns the byte stream sent by the client, but no identifying in-
formation about the client.
14. AWeb browser will cache objects locally in its browser cache. An institution
might also have a Web cache, which each browser contacts to satisfy its Web
requests. If the requested Web page is not in the Web cache, the Web cache will
request and receive the object from the origin server, cache the object locally (in
case some other browser requests the object), and return the object to the re-
quest browser. When the local DNS gets a translation request for a name that it
is not in its cache, it will obtain the name/address translation pair and then
cache this value so that future requests for translating the same name can be sat-
isfied by the local DNS server, without going out to the larger DNS system.
15. The MAIL FROM: in SMTP is a message from the SMTP client that identi-
fies the sender of the mail message to the SMTP server. The From: on the
mail message itself is NOT an SMTP message, but rather is just a line in the
body of the mail message.
16. FM radio, broadcast TV, and newspapers are all push systems. Information is
sent out via these media regardless of whether or not anyone is tuning in to
the TV or radio stations, or buying the newspaper. Also, content is not indi-
vidualized for the receiver. The Web is a pull system because data is not
transmitted to the browser unless it is explicitly requested by the browser.
Some automated phone response systems are non-Internet pull systems. For
example, I may call a number to get my horoscope, and enter my birth date
in order to have my horoscope read to me.
17. a. The advantage of sending the QueryHit message directly over a TCP
connection from Bob to Alice is that the QueryHit message is routed by
the underlying Internet without passing through intermediate peers; thus,
the delay in sending the message from Bob to Alice should be substan-
tially less. The disadvantage is that each peer that has a match would ask
Alice to open a TCP connection; Alice may therefore have to opens tens
or hundreds of TCP connections for a given query.
b. When a QueryHit message enters a peer, the peer records the MessageID
in a table along with an identifier of the TCP socket from which the mes-
sage arrived. When the same peer receives a QueryHit message with the
same MessageID, it indexes the table and determines the socket to which
it should forward the message.
CH01-02_p1-30 6/15/06 4:34 PM Page 29

c. As the query message propagates through peers, each peer includes its IP
address in the query message, creating a list of IP addresses. When there
is a query match, the peer with the match includes the list in the Query-
Hit message. When a peer receives a QueryHit, it finds the preceding
node in the list and forwards the message to that preceding peer.
18. The welcoming socket is used in a server to wait for incoming client connec-
tion requests in connection-oriented (TCP-based) communication. A new
socket is created on return from the accept(). This new socket is then used by
the server to communicate back to the client.
19. If all traffic is UDP, then the number of sockets will equal the number of
ports. If there is TCP traffic, then the number of sockets will be greater, be-
cause each welcoming socket, and the socket created as a result of
accept()ing a connection will have the same destination port number.
20. See page 155 of your textbook.
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